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s344n2d4d5 [400]
3 years ago
15

What is the square root of 124

Mathematics
1 answer:
marysya [2.9K]3 years ago
5 0

Answer:

  √124 simplifies to 2√31 ≈ 11.135529

Step-by-step explanation:

√124 = √(4·31) = (√4)(√31) = 2√31

A calculator can tell you the numerical value. It is an irrational number near ...

  11.135528725660043844...

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A≈143.17

Step-by-step explanation:

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Every Tuesday, a bake shop offers a $5 special for 5 items. A customer can get any combination of bagels and donuts, but no more
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6 0
3 years ago
Help please I really need it
gtnhenbr [62]
So we want to solve <span><span><span>2x </span>− 40 </span>= <span>3y </span></span>for x<span><span><span><span>2x </span>− 40 </span>+ 40 </span>= <span><span>3y </span>+ 40</span></span>(Add 40 to both sides!)<span><span>2x </span>= <span><span>3y </span>+ <span>40
Now divide both sides by 2!
</span></span></span>\frac{2x}{2} = \frac{3y + 40}{2} \ \textgreater \  x =  \frac{3}{2}y + 20
Now substitute \frac{3}{2}y + 20 for x in 2y = -4x + 48&#10;&#10;
2y = -4( \frac{3}{2}y + 20) + 48
Which then gives us..
<span><span>2y </span>= <span><span>−<span>6y </span></span>− 32</span></span>(Make sure to simplify both sides of the equation)<span><span><span>2y </span>+ <span>6y </span></span>= <span><span><span>−<span>6y </span></span>− 32 </span>+ <span>6y</span></span></span>(Now add 6y to both sides)<span><span>8y </span>= <span>−<span>32
Now divide both sides by 8
</span></span></span>\frac{8y}{8} = \frac{-32}{8} \ \textgreater \  y = -4
Now we have to substitute -4 for y in x =  \frac{3}{2}y + 20 \ \textgreater \   \frac{3}{2}(-4) + 20 \ \textgreater \  x = 14
So we end up getting...
<span>x = <span><span>14<span> and </span></span>y </span></span>= <span>−<span>4.
</span></span>Hope this helps!
7 0
3 years ago
What is the sum of the series infinity e n-1 -7(3/8)^n?
soldier1979 [14.2K]

Let

S_N = \displaystyle \sum_{n=1}^N -7 \left(\frac38\right)^n = -7\left(\dfrac38 + \dfrac{3^2}{8^2} + \dfrac{3^3}{8^3} + \cdots + \dfrac{3^N}{8^N}\right)

Then

\dfrac38 S_N = -7\left(\dfrac{3^2}{8^2} + \dfrac{3^3}{8^3} + \dfrac{3^4}{8^4} + \cdots + \dfrac{3^{N+1}}{8^{N+1}}\right)

S_N - \dfrac38 S_N = -7 \left(\dfrac38 - \dfrac{3^{N+1}}{8^{N+1}}\right)

\dfrac58 S_N = -\dfrac{21}8 \left(1 - \left(\dfrac38\right)^N\right)

S_N = -\dfrac{21}5 \left(1 - \left(\dfrac38\right)^N\right)

As N\to\infty, the exponential term will converge to zero, so the infinite sum converges to

\displaystyle \sum_{n=1}^\infty -7 \left(\frac38\right)^n = \lim_{n\to\infty} S_N = \boxed{-\dfrac{21}5}

6 0
2 years ago
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